320 XV NORMED ALGEBRAS AND SPECTRAL THEORY For each x e A, the mapping #f~»(^*)(x) = %(x) is continuous on X(A), by the definition of the weak topology. Moreover, we have and therefore 9(xy) = (9x)(9y). The relation ye = 1 is obvious. The equality \<9x\ = p(;c) will follow from (ii) and the definition of p(x) (15.2.7). Consider therefore the first assertion in (ii). We know already that #(*) 6 Sp(jc) for all ;ce A and x e X(A) (15.3.1). Let us show conversely that, for each /I e Sp(x), there exists a character x such that x(x) = A, or equivalently that x(x — Xe) = 0. Since x — he is not invertible, the ideal A(x — he) generated by this element is distinct from A. In view of (1 5.3.1 (iii)) we are therefore reduced to proving the following lemma: (15.3.4.1) In a separable commutative Banach algebra A with unit element e 7^ 0, every ideal a ^ A is contained in a maximal ideal. Let (xn) be a dense sequence in A. We define inductively an increasing sequence (an) of closed ideals ^A (n ^ 0) as follows. Take a0 == a (which is distinct from A, by (15.3.1.1)). In the quotient Banach algebra A/ an.l con- sider the image xn ofxn and an element Art of its spectrum (which by (15.2.4) is not empty), so that xn — lne (where e is the canonical image of e in A/an-i) is not invertible in A/aw-t. It follows that the ideal a'n in A generated by an_! and xn — ln e is distinct from A, and therefore an = a^ is distinct from A. The union m' of the ideals an is therefore also an ideal ^ A, and hence the closure m of m' is also an ideal ^ A. We shall show that m is maximal, which will complete the proof. If e" is the unit element of A/m, and x"n the image of xn in A/m, then by construction we have x"n = kne"\ in other words, all the x£ belong to the closed subalgebra (5.9.2) Ce" of A" = A/m. Since they form a dense subset of A" (3.11.4), it follows that A" = O", and there- fore m is maximal. Remarks (15.3.5) It should be noted that the Gelfand transformation x^&x is not necessarily injective. For example, if x is nilpotent but not zero, say of = 0, then the image of Sp(x) under £\-> C* is {0} ((15.2.3.1) and (15.2.1)) and therefore Sp(x) = {0}, hence %(x) = 0 for all characters %, although x ^ 0. This example also shows that \\9x\\ need not be equal to ||x||. But even if ^ is isometric (in which case the image 9(A) is closed'm #C(X(A)) (3.14.4)), may not be equal to «c(*(A)) (15.3.8).nt that